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How many five-digit numbers can be formed using the digits

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Intern
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How many five-digit numbers can be formed using the digits [#permalink] New post 11 Oct 2008, 14:09
How many five-digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?

* 15
* 96
* 120
* 181
* 216

Can someone explain solution approach for this question.
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Re: m4 - 32 - [#permalink] New post 11 Oct 2008, 20:08
IMO E?

lets take first digit as 5.. so remaining 4 digit can be arrange in this pattern

pattern1: 5 0 4 2 1 ...sum =12 /3 ..okay..so total possible combination 1*4!= 24

pattern1: 5 4 3 2 1 ...sum =15 /3 ..okay..so total possible combination 1*4!= 24

so for if first digit is 5 we have 48 numbers

lets take first digit as 4.. so remaining 4 digit can be arrange in this pattern

pattern1: 4 0 5 2 1 ...sum =12 /3 ..okay..so total possible combination 1*4!= 24

pattern1: 4 5 3 2 1 ...sum =15 /3 ..okay..so total possible combination 1*4!= 24

so for if first digit is 4 we have 48 numbers

same goes on for all the number and you willend up with 48*5 =240..which is not in answer

so bestnear by answer is 216 :)
* i will follow this proc.when i have very less time say 90sec. in exam and i have to mark some answer so i will go for E. i knowi madesomesmallmistake somewhere..please rectify it
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Re: m4 - 32 - [#permalink] New post 11 Oct 2008, 20:48
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lalmanistl wrote:
How many five-digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?

* 15
* 96
* 120
* 181
* 216

Can someone explain solution approach for this question.


5 digit integers made of "1, 2, 3, 4, and 5" & "0, 1, 2, 4, and 5" are divisible by 3.
so the no. of possibilities are = 5! + 4x4! = 120 + 96 = 216
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Re: m4 - 32 - [#permalink] New post 12 Oct 2008, 03:13
5 digit number formed which is divisible by 3 can be formed by combonation of

1,2,3,4&5 in 5! ways
OR
0,1,2,4 & 5 with 0 will not be Most Significant digit in 4X4!

Total ways = 216
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Re: m4 - 32 - [#permalink] New post 12 Oct 2008, 06:27
Only when we are using the set of numbers 0,1,2,4 & 5 or 1,2,3,4 & 5 should the 5 digit number be divisible by 3

with the first set if we remove the 5 digits set that will start with 0 that is 24 from 5! for the whole probability then the difference will be
5! - 4! = 120 - 24 = 96

then we add the number of probabilities for the second set

therefore we get the result of

5! + 5! - 4! = 120 + 120 -24 = 240 - 24 = 216

thanks

Mohamed
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Re: m4 - 32 - [#permalink] New post 27 Jan 2009, 19:55
GMAT TIGER wrote:
lalmanistl wrote:
How many five-digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?

* 15
* 96
* 120
* 181
* 216

Can someone explain solution approach for this question.


5 digit integers made of "1, 2, 3, 4, and 5" & "0, 1, 2, 4, and 5" are divisible by 3.
so the no. of possibilities are = 5! + 4x4! = 120 + 96 = 216


Can someone tell me why we multiply 5*4*3*2*1? and why we multiply 4*3*2*1? is there a rule for using factorial in creating number combinations?
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Re: m4 - 32 - [#permalink] New post 28 Jan 2009, 09:45
We have 5 ways to fill up the first place:5
5 ways to fill up the second :5
4 ways to fill up the thrid:4
3 ways to fill up the fourth:3 and 2 ways to fill up the fifth.
Multiplying, I get 600. What is wrong with this approach?


x-ALI-x wrote:
GMAT TIGER wrote:
lalmanistl wrote:
How many five-digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?

* 15
* 96
* 120
* 181
* 216

Can someone explain solution approach for this question.


5 digit integers made of "1, 2, 3, 4, and 5" & "0, 1, 2, 4, and 5" are divisible by 3.
so the no. of possibilities are = 5! + 4x4! = 120 + 96 = 216


Can someone tell me why we multiply 5*4*3*2*1? and why we multiply 4*3*2*1? is there a rule for using factorial in creating number combinations?

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tusharvk

Re: m4 - 32 -   [#permalink] 28 Jan 2009, 09:45
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